How do you simplify #(4s + st) /( 3s)#?

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Answer 1 · 2016-09-07T13:21:43

#(1) / (3) (4 + t)#

We have: #(4 s + s t) / (3 s)#

First, let's factorise the numerator by taking #s# in common:

#= (s (4 + t)) / (3 s)#

We can now cancel the #s# terms:

#= (4 + t) / (3)#

#= (4) / (3) + (1) / (3) t#

#= (1) / (3) (4 + t)#